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Thread: "problem of inequalities"

  1. #1
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    "problem of inequalities"

    "problem of inequalities"
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  2. #2
    GAMMA Mathematics
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    Please post only one question at a time.

    I will help you solve the first problem.

    \sqrt{3x+1}<x+1

    Square both sides:

    3x+1<(x+1)^2 =\Rightarrow 3x+1<x^2+2x+1

    0<x^2-x

    0<x(x-1)

    0>x, 1<x

    Only one solution will work in this case

    1 < x \Rightarrow x > 1

    EDIT: DivideBy0 pointed out a spot that wouldn't work, which I suspected... grr

    3x+1 \ge 0

    x \ge \frac{-1}{3}
    Last edited by colby2152; Jan 4th 2008 at 05:23 AM.
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  3. #3
    Senior Member DivideBy0's Avatar
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    I think you made a little typo there which led to a mistake.

    x<0 or x > 1

    But, to be defined,

    3x+1\geq 0

    3x\geq -1

    x \geq -\frac{1}{3}

    So the solution set is \{x:-\frac{1}{3}\leq x < 0 \cup x>1\}
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by afeasfaerw23231233 View Post
    "problem of inequalities"
    For the second question, look at the graph of that function.

    Can you see it intercepts the axis only once?

    So  x \geq 3 satisfies the equation
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    Last edited by janvdl; Jan 4th 2008 at 05:22 AM. Reason: Forgot \geq
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  5. #5
    Senior Member DivideBy0's Avatar
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    \frac{x-2}{x+2} > \frac{x^2-4}{x^2+4}

    \left(\frac{x-2}{x+2}\right)\left(\frac{x-2}{x-2}\right)>\frac{x^2-4}{x^2+4}

    \frac{(x-2)^2}{x^2-4}>\frac{x^2-4}{x^2+4}

    \left(\frac{(x-2)^2}{x^2-4}\right)\left(\frac{x^2+4}{x^2+4}\right)>\frac{x^  2-4}{x^2+4}\left(\frac{x^2-4}{x^2-4}\right)

    \frac{(x^2+4)(x-2)^2}{x^4-16}>\frac{(x^2-4)^2}{x^4-16}

    Case 1

    If x^4-16>0\Rightarrow x > 2 or x<-2,

    (x^2+4)(x-2)^2>(x^2-4)^2

    x^4-4x^3+8x^2-16x+16>x^4-8x^2+16

    4x^3-16x^2+16x<0

    4x(x^2-4x+4)<0

    4x(x-2)^2<0

    Which is true so long as x<0.

    Case 2

    If x^4-16<0\Rightarrow -2 < x< 2, after some simplification we arrive at

    4x(x-2)^2>0

    Which is true so long as x>0.



    So the two regions where the inequality is true is when

    -2<x<2 AND x>0 \Rightarrow 0<x<2.

    Or

    x<-2 \cap x>2 AND x<0 \Rightarrow x<-2.

    So the possible values of x are \{x:0<x<2 \cup x<-2\}
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